JEE Main 2015 · 4 Apr · Shift 1 · Mathematics Q83

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MathematicsSingle correct+4 / -1Question 83 of 90

The distance of the point (1, 0, 2) from the point of intersection of the line $\dfrac{x-2}3=\dfrac{y+1}4=\dfrac{z-2}{12}$ and the plane $x-y+z=16$ , is :

2\sqrt{14}

3\sqrt{21}

Answer and worked solution

Official answer: D

The line is $(2+3t,\,-1+4t,\,2+12t)$ . In the plane: $(2+3t)-(-1+4t)+(2+12t)=16$ gives $t=1$ , so the point is (5, 3, 14).
$\text{Distance from }(1,0,2)=\sqrt{16+9+144}=13.$

Topics:Equation of a line in space Equation of a plane

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JEE Main 2015 · 4 Apr · Shift 1, Mathematics question 83